Judy Zahrndt wants to name her baby so that his monogram (first, middle, and last initials) will be distinct letters in alphabetical order and he will share her last name. How many different monograms could she select?
The last initial must be Z. The other two letters can be picked in C(25,2) ways. For any such pair, there is only one order that they can occur in the monogram, so there are C(25,2) = 25*24/2 = 300 ways.
word problem permutations help?
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May 1st, 2010 at 2:02 pm
C(25,2) = 300
Another way is to think of it like this: There are 25 choices for the first letter and 24 for the second and there are 2! ways to arrange the first and middle names. So it is 25*24/2! = 300
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May 1st, 2010 at 2:18 pm
The last initial must be Z. The other two letters can be picked in C(25,2) ways. For any such pair, there is only one order that they can occur in the monogram, so there are C(25,2) = 25*24/2 = 300 ways.
References :
May 1st, 2010 at 3:05 pm
Hi Bill
Let’s look at it logically
Z as the last name initial is fixed
So we can have
A as the first name initial and any of the remaining 24 letters between A & Z (both excluded)as the middle initial (24 combinations) OR
B as the first name initial and any of the remaining 23 letters between B & Z (both excluded)as the middle initial (23 combinations) OR
C as the …. and so on till
X as the first name initial and any of the remaining 1 letter between X & Z (both excluded)as the middle initial(1 combination).
OR in permutation and combination means addition
So it boils down to
24 + 23 + 22 +……..+ 1
sum of n natural numbers = n(n+1)/2
24(24+1)/2 = 12*25 = 300
Judy can select 300 different monograms
References :